C3 June 2016 Q9
9. The amount of an antibiotic in the bloodstream, from a given dose, is modelled by the formula\[x = D\mathrm{e}^{-0.2t}\]where \(x\) is the amount of the antibiotic in the bloodstream in milligrams, \(D\) is the dose given in milligrams and \(t\) is the time in hours after the antibiotic has been given.
A first dose of 15 mg of the antibiotic is given.
A second dose of 15 mg is given 5 hours after the first dose has been given. Using the same model for the second dose,
No more doses of the antibiotic are given. At time \(T\) hours after the second dose is given, the total amount of the antibiotic in the bloodstream is 7.5 mg.
| Scheme | Marks |
|---|---|
| Subs \(D = 15\) and \(t = 4\) \(x = 15\mathrm{e}^{-0.2 \times 4} = 6.740\ (mg)\) | M1A1 |
| (2) |
Notes
M1: Attempts to substitute both \(D = 15\) and \(t = 4\) in \(x = D\mathrm{e}^{-0.2t}\)
It can be implied by sight of \(15\mathrm{e}^{-0.8}\), \(15\mathrm{e}^{-0.2 \times 4}\) or awrt 6.7
Condone slips on the power. Eg you may see -0.02
A1: CAO 6.740 (mg) Note that 6.74 (mg) is A0
| Scheme | Marks |
|---|---|
| \(15\mathrm{e}^{-0.2 \times 7} + 15\mathrm{e}^{-0.2 \times 2} = 13.754(mg)\) | M1A1* |
| (2) |
Notes
M1: Attempt to find the sum of two expressions with \(D\) =15 in both terms with \(t\) values of 2 and 7
Evidence would be \(15\mathrm{e}^{-0.2 \times 7} + 15\mathrm{e}^{-0.2 \times 2}\) or similar expressions such as \(\left(15\mathrm{e}^{-1} + 15\right)\mathrm{e}^{-0.2 \times 2}\)
Award for the sight of the two numbers awrt 3.70 and awrt 10.05, followed by their total awrt 13.75
Alternatively finds the amount after 5 hours, \(15\mathrm{e}^{-1} =\) awrt 5.52 adds the second dose = 15 to get a total of awrt 20.52 then multiplies this by \(\mathrm{e}^{-0.4}\) to get awrt 13.75.
Sight of 5.52+15=20.52\(\to\) 13.75 is fine.
A1*: cso so both the expression \(15\mathrm{e}^{-0.2 \times 7} + 15\mathrm{e}^{-0.2 \times 2}\) and \(13.754(mg)\) are required
Alternatively both the expression \(\left(15\mathrm{e}^{-0.2 \times 5} + 15\right) \times \mathrm{e}^{-0.2 \times 2}\) and \(13.754(mg)\) are required.
Sight of just the numbers is not enough for the A1*
| Scheme | Marks |
|---|---|
| \(15\mathrm{e}^{-0.2 \times T} + 15\mathrm{e}^{-0.2 \times (T + 5)} = 7.5\) | M1 |
| \(15\mathrm{e}^{-0.2 \times T} + 15\mathrm{e}^{-0.2 \times T}\mathrm{e}^{-1} = 7.5\) \(15\mathrm{e}^{-0.2 \times T}(1 + \mathrm{e}^{-1}) = 7.5 \Rightarrow \mathrm{e}^{-0.2 \times T} = \dfrac{7.5}{15(1 + \mathrm{e}^{-1})}\) | dM1 |
| \(\mathrm{T} = -5\ln\left(\dfrac{7.5}{15\left(1 + \mathrm{e}^{-1}\right)}\right) = 5\ln\left(2 + \dfrac{2}{\mathrm{e}}\right)\) | A1, A1 |
| (4) | |
| (8 marks) |
Notes
M1: Attempts to write down a correct equation involving \(T\) or \(t\). Accept with or without correct bracketing
Eg. accept \(15\mathrm{e}^{-0.2 \times T} + 15\mathrm{e}^{-0.2 \times (T \pm 5)} = 7.5\) or similar equations \(\left(15\mathrm{e}^{-1} + 15\right)\mathrm{e}^{-0.2 \times T} = 7.5\)
dM1: Attempts to solve their equation, dependent upon the previous mark, by proceeding to \(\mathrm{e}^{-0.2 \times T} = \ldots\)
An attempt should involve an attempt at the index law \(x^{m+n} = x^m \times x^n\) and taking out a factor of \(\mathrm{e}^{-0.2 \times T}\) Also score for candidates who make \(\mathrm{e}^{+0.2 \times T}\) the subject using the same criteria
A1: Any correct form of the answer, for example, \(-5\ln\left(\dfrac{7.5}{15\left(1 + \mathrm{e}^{-1}\right)}\right)\)
A1: CSO \(\mathrm{T} = 5\ln\left(2 + \dfrac{2}{\mathrm{e}}\right)\) Condone \(t\) appearing for \(T\) throughout this question.
Alt (c) using lns
| Scheme | Marks |
|---|---|
| \(15\mathrm{e}^{-0.2 \times T} + 15\mathrm{e}^{-0.2 \times (T + 5)} = 7.5\) | M1 |
| \(15\mathrm{e}^{-0.2 \times T} + 15\mathrm{e}^{-0.2 \times T}\mathrm{e}^{-1} = 7.5\) \(\mathrm{e}^{-0.2 \times T}(1 + \mathrm{e}^{-1}) = 0.5 \Rightarrow -0.2 \times T + \ln(1 + \mathrm{e}^{-1}) = \ln 0.5\) | dM1 |
| \(\Rightarrow T = \dfrac{\ln 0.5 - \ln(1 + \mathrm{e}^{-1})}{-0.2}, \Rightarrow T = 5\ln\left(2 + \dfrac{2}{\mathrm{e}}\right)\) | A1, A1 |
| (4) |
You may see numerical attempts at part (c).
Such an attempt can score a maximum of two marks.
This can be achieved either by
Method One
1st Mark (Method): \(15\mathrm{e}^{-0.2 \times T} + \text{awrt } 5.52\mathrm{e}^{-0.2 \times T} = 7.5 \Rightarrow \mathrm{e}^{-0.2 \times T} = \text{awrt } 0.37\)
2nd Mark (Accuracy): T=-5ln(awrt 0.37) or awrt 5.03 or \(\mathrm{T} = -5\ln\left(\dfrac{7.5}{\text{awrt } 20.52}\right)\)
Method Two
1st Mark (Method ): \(13.754\mathrm{e}^{-0.2 \times T} = 7.5 \Rightarrow T = -5\ln\left(\dfrac{7.5}{13.754}\right)\) or equivalent such as 3.03
2nd Mark (Accuracy): 3.03+2=5.03 Allow \(-5\ln\left(\dfrac{7.5}{13.754}\right) + 2\)
Method Three (by trial and improvement)
1st Mark (Method): \(15\mathrm{e}^{-0.2 \times 5} + 15\mathrm{e}^{-0.2 \times 10} = 7.55\) or \(15\mathrm{e}^{-0.2 \times 5.1} + 15\mathrm{e}^{-0.2 \times 10.1} = 7.40\) or any value between
2nd Mark (Accuracy): Answer \(T\) =5.03.